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Integers k such that floor(sqrt(k)) + floor(sqrt(k/2)) divides k.
4

%I #17 Mar 08 2019 23:50:56

%S 1,2,6,8,10,15,21,40,60,65,90,102,119,126,133,160,168,176,216,225,270,

%T 290,319,330,341,384,396,408,468,546,615,630,704,736,782,799,816,918,

%U 1007,1026,1120,1160,1218,1239,1260,1342,1364,1386,1495,1632,1750,1775

%N Integers k such that floor(sqrt(k)) + floor(sqrt(k/2)) divides k.

%C This sequence is infinite. Proof: if x > y > 1 satisfies x^2 - 2*y^2 = -1 (x=A002315(j), y=A001653(j+1), j>0), then x < 2*y. Let k = 2*y^2 + m; then 0 <= m <= 2*x - 1, because x^2 < x^2 + my + 1 < (x+1)^2 and y^2 <= y^2 + m/2 < y^2 + 2*y, floor(sqrt(k)) = floor(sqrt(x^2+m+1)) = x and floor(sqrt(k/5)) = floor(sqrt(y^2+m/2)) = y. x + y < 2*x, so by the pigeonhole principle there exists a number m belonging to [0, 2*x - 1] such that x + y | 2*y^2 + m, so such k is a term.

%o (PARI) is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/2))) == 0;

%Y Cf. A001653, A002315.

%K nonn

%O 1,2

%A _Jinyuan Wang_, Feb 23 2019